|a−b|²=|a|²+|b|²−2a·b = 1+1−2(1/2)=1. So |a−b|=1. Wait—correct value is 1, not 1/√2.
If |a|=|b|=1 and a·b=1/2, then |a−b| equals
A 1
B √2
C 1/√2
D √3
Use |a−b|²=|a|²+|b|²−2a·b. Substituting gives 1+1−2(1/2)=1, so |a−b|=1. This matches a 60° angle case.
For |a|=3, |b|=4 and a·b=6, the angle between them is
A 30°
B 60°
C 45°
D 90°
cosθ=(a·b)/(|a||b|)=6/(3×4)=6/12=1/2, so θ=60°. Dot product directly links magnitudes and included angle.
If vectors a,b satisfy |a+b|=|a−b|, then
A a∥b
B |a|=|b|
C a=b
D a⊥b
Square both: |a+b|²=|a|²+|b|²+2a·b and |a−b|²=|a|²+|b|²−2a·b. Equality gives a·b=0, so vectors are perpendicular.
If a×b = b×c = c×a and a×b ≠ 0, then
A a+b+c=0
B a=b=c
C a ⟂ b ⟂ c
D |a| all equal
From c = −(a+b), we get b×c = b×(−a−b)=a×b and c×a = (−a−b)×a = a×b. Nonzero cross product avoids the “all zero” parallel case.
If a×b = b×c = c×a, then vectors are
A all perpendicular
B equal magnitudes
C all parallel
D coplanar only
a×b=b×c implies b×(a−c)=0, so a−c is parallel to b. Similarly other relations force all three to share one direction, hence all are parallel (or some zero).
If a·(b×c)=6 and b·(c×a)=6, then
A cyclic property
B always true
C sign changes
D perpendicular vectors
Scalar triple product is cyclic: a·(b×c)=b·(c×a)=c·(a×b). Same value appears under cyclic shift, so equality is expected without extra conditions.
If a·(b×c)=6, then c·(a×b) equals
A −6
B 0
C 12
D 6
Cyclic permutation keeps scalar triple product unchanged. Since a·(b×c)=6, then b·(c×a)=6 and c·(a×b)=6 as well.
If a,b,c are nonzero and a×b = a×c, then
A b=c
B b ⟂ c
C b−c ∥ a
D a ⟂ (b−c)
a×b=a×c ⇒ a×(b−c)=0. For nonzero a, cross product zero means (b−c) is parallel to a (or b=c as a special case).
If a·b = |a||b|cosθ, then a·(−b) equals
A −|a||b|cosθ
B +|a||b|cosθ
C |a||b|sinθ
D 0 always
Replacing b by −b reverses direction, so dot product changes sign: a·(−b)=−(a·b). Magnitudes stay same, only direction flips.
If a×(b×c)=0 and b×c ≠ 0, then a is
A ⟂ b
B ∥ (b×c)
C ⟂ c
D coplanar with b,c
a×(b×c)=0 means a is parallel to (b×c) or a is zero. Given b×c≠0, the direction exists, so a must be parallel to that vector.
If a×(b×c)=b(a·c)−c(a·b), this identity is
A Lagrange identity
B scalar triple rule
C BAC–CAB identity
D section formula
The vector triple product identity a×(b×c)=(a·c)b−(a·b)c is known as BAC–CAB. It converts nested cross product to dot products.
If a·b=0 and a·c=0, then a is parallel to
A b×c
B b+c
C b−c
D b·c
If a is perpendicular to both b and c, then a is along the normal to the plane of b and c. That normal direction is given by b×c (if b and c not parallel).
If b×c=0, then statement true is
A b ⟂ c
B b·c=0
C |b|=|c|
D b ∥ c
Cross product zero means sinθ=0, so θ=0° or 180°, indicating vectors are parallel or anti-parallel. Magnitudes need not be equal.
If |a|=2, |b|=3, and |a×b|=3, then sinθ equals
A 1/3
B 1/2
C 2/3
D 3/2
|a×b|=|a||b|sinθ ⇒ 3=2×3×sinθ ⇒ sinθ=3/6=1/2. Angle could be 30° or 150°.
|a×b|=|a||b||sinθ| and |a·b|=|a||b||cosθ|. Equality gives |sinθ|=|cosθ| ⇒ θ=45° or 135°. Standard answer is 45°.
If a=(1,1,1) and b=(1,2,3), then vector projection of b on a is
A (6)a/3
B (6)a/3?
C (2)a/3
D (2)a
proj of b on a = ((b·a)/(a·a))a. Here b·a=1+2+3=6 and a·a=3. So projection vector = (6/3)a=2a. Options mismatch; replace correctly.
If a=(1,1,1) and b=(1,2,3), projection of b on a is
A a/2
B 3a
C 6a
D 2a
Vector projection is ((b·a)/(a·a))a. Compute b·a=6, a·a=3, so factor is 2. Hence projection vector equals 2a = (2,2,2).
For points A(1,0,0) and B(0,1,0), internal division in 1:2 gives P as
A (2/3,1/3,0)
B (1/3,2/3,0)
C (1/2,1/2,0)
D (−1/3,2/3,0)
For internal ratio AP:PB=m:n, P=(na+mb)/(m+n). Here m=1,n=2, so P=(2A+1B)/3=(2/3,1/3,0), closer to A since AP is smaller part.
If points A,B,C have position vectors a,b,c, then AB and AC are perpendicular if
A (b+a)·(c+a)=0
B (b−a)·(c−a)=0
C (b−c)·a=0
D (a+b+c)=0
Vectors AB=b−a and AC=c−a. For perpendicularity, their dot product must be zero: (b−a)·(c−a)=0. This is a direct vector condition.
If A,B,C are collinear, then (b−a) and (c−a) are
A perpendicular
B equal always
C parallel
D unit vectors
Collinearity means AB and AC lie along same line direction. Hence AB is a scalar multiple of AC, so they are parallel, giving cross product zero.
If a,b,c are position vectors of triangle vertices, centroid is
A (a+b+c)/3
B (a+b)/2
C (a−b+c)/3
D (a+b−c)/3
Centroid is average of vertex position vectors. It divides each median in 2:1 ratio from vertex and lies inside the triangle for non-collinear vertices.
If G is centroid of triangle ABC, then vector AG equals
A (a+b−2c)/3
B (b+c−2a)/3
C (a+c−2b)/3
D (a+b+c)/3
G has position vector (a+b+c)/3. So AG = OG−OA = (a+b+c)/3 − a = (b+c−2a)/3. This is a useful centroid relation.
If D is midpoint of BC, then AD vector equals
A (b+c−2a)/2
B (a+b)/2 − c
C (a+c)/2 − b
D (b+c)/2 − a
Midpoint D has vector (b+c)/2. Then AD = OD−OA = (b+c)/2 − a. This expresses median direction from vertex A.
If a,b are nonzero and a·b = |a||b|, then a and b are
A perpendicular
B parallel opposite
C parallel same
D skew vectors
a·b=|a||b|cosθ. Equality with |a||b| means cosθ=1 so θ=0°. Hence they are in same direction (parallel).
If |a+b|=0, then
A a=−b
B a=b
C a⊥b
D |a|=0 only
If a+b=0 vector, then b=−a. That means equal magnitudes and opposite directions. This is a strong condition, not just perpendicularity.
If a·(b×c)=0 and b×c≠0, then a is
A parallel to b
B in plane of b,c
C parallel to c
D perpendicular to plane
b×c is normal to plane of b and c. If a·(b×c)=0, then a is perpendicular to that normal, meaning a lies in the plane of b and c.
If scalar triple product is nonzero, vectors are
A coplanar
B linearly dependent
C not coplanar
D parallel pair
Nonzero a·(b×c) means volume is nonzero, so vectors span 3D space and are not coplanar. They are linearly independent.
If b×c is perpendicular to a, then a·(b×c) is
A 0
B 1
C negative
D maximum
Dot product of perpendicular vectors is zero. Since a is perpendicular to (b×c), scalar triple product becomes zero, meaning no volume component along a.
For direction cosines (l,m,n), possible set is
A (1,1,0)
B (1/2,1/2,1/2)
C (2,0,0)
D (1/√2,1/√2,0)
Direction cosines must satisfy l²+m²+n²=1. For option C: 1/2+1/2+0=1, valid. Others do not satisfy this condition.
If line direction ratios are (2,−1,2), a perpendicular plane normal can be
A (2,−1,2)
B (1,0,−1)
C (0,1,0)
D (1,1,1)
A plane perpendicular to a given line has normal vector parallel to the line’s direction vector. So the plane’s normal can be same as direction ratios (2,−1,2).
If plane normal is (2,−1,2), plane equation through origin is
A 2x+y+2z=0
B x−2y+z=0
C 2x−y+2z=0
D 2x−y−2z=0
Plane through origin with normal n satisfies r·n=0. With n=(2,−1,2), we get 2x−y+2z=0. This is direct from dot product.
Magnitude is √(2²+5²+(−4)²)=√(4+25+16)=√45. This equals area of parallelogram formed by vectors a and b.
If |a×b|=√45, triangle area is
A √45/2
B √45
C 45/2
D √90
Triangle area with sides a and b from same vertex is half the parallelogram area. So area = |a×b|/2 = √45/2.
If a,b,c are such that a = b×c, then a is
A in plane b,c
B parallel to b
C perpendicular to b,c
D parallel to c
Cross product b×c is perpendicular to both b and c. So if a equals that cross product, a is normal to the plane containing b and c.
If a·b = 0 and |a|=|b|, then |a+b| equals
A |a|
B √2|a|
C 2|a|
D 0
Using |a+b|²=|a|²+|b|²+2a·b, with a·b=0 and equal magnitudes gives 2|a|². So |a+b|=√2|a|.
If |a+b|=|a|+|b|, then vectors are
A parallel same
B perpendicular
C parallel opposite
D coplanar only
Equality in triangle inequality occurs only when vectors are in the same direction. So a and b must be parallel with angle 0°, giving maximum possible resultant length.
If |a+b|=||a|−|b||, then vectors are
A parallel same
B perpendicular
C parallel opposite
D skew only
Minimum resultant occurs when vectors are opposite in direction (θ=180°). Then magnitude becomes difference of magnitudes: |a+b| = ||a|−|b||.
If a=(2,1,0), b=(1,−1,0), then angle is 90° because
A a×b=0
B |a|=|b|
C a=b
D a·b=0
Dot product a·b = 2·1 + 1·(−1) + 0·0 = 1 ≠ 0. So not 90°. Replace with correct hard check:
If a=(1,1,0), b=(1,−1,0), then vectors are perpendicular since
A a×b=0
B a·b=0
C |a|=|b|
D a=b
Compute a·b = 1·1 + 1·(−1) + 0·0 = 0. Dot product zero confirms 90° angle between nonzero vectors.
If a=(1,1,0), b=(1,−1,0), then |a×b| equals
A 2
B 0
C 1
D √2
Cross product is (0,0, 1·(−1) − 1·1) = (0,0,−2). Magnitude is 2. This equals parallelogram area in xy-plane.
For three points A(a),B(b),C(c), area of triangle is
A |(b−a)×(c−a)|
B |(b−a)·(c−a)|/2
C |(b−a)×(c−a)|/2
D |(a+b+c)|/3
Using vectors AB=b−a and AC=c−a, parallelogram area is |AB×AC|. Triangle is half of that, so area = |(b−a)×(c−a)|/2.
Maximum magnitude of scalar triple product is |a||b||c|, achieved when vectors are mutually perpendicular and appropriately oriented, giving a rectangular parallelepiped.
If a·(b×c)=0 and a is not in plane of b,c, then
A b×c=0
B impossible
C a=b+c
D |a|=0
If b×c≠0, then a·(b×c)=0 forces a to lie in plane of b and c. So if a is not in that plane, the only way is b×c=0 (coplanar degenerate).
If direction cosines are (l,m,n) and l=0, then line is
A parallel x-axis
B along x-axis
C perpendicular x-axis
D skew to x-axis
l=cosα where α is angle with x-axis. If l=0, α=90°, meaning the line is perpendicular to x-axis. It lies entirely in the yz-directional plane.
If l=m=0, then direction is
A along z-axis
B along y-axis
C along x-axis
D in xy-plane
l=m=0 implies n²=1, so n=±1. That means direction is purely along z-axis, either +k or −k direction.
If a=(1,2,3) and b=(4,5,6), then scalar triple with c=(7,8,9) is
A 3
B 9
C 0
D 12
These vectors are linearly dependent because c−b = b−a = (3,3,3). So they lie in same plane and scalar triple product becomes zero, giving zero volume